Answer
$$\frac{1}{4}\left( {-1 - 2{x^2}} \right)\cos 2x + \frac{1}{2}x\sin 2x + C$$
Work Step by Step
$$\eqalign{ & \int {{x^2}\sin 2xdx} \cr & {\text{Let }}u = {x^2},\,\,\,\,dv = \sin 2xdx \cr & \,\,\,\,\,\,du = 2xdx\,\,\,\,\,v = - \frac{1}{2}\cos 2x \cr & {\text{Using the integration by parts formula}} \cr & \int {{x^2}\sin 2xdx} = \left( {{x^2}} \right)\left( { - \frac{1}{2}\cos 2x} \right) - \int {\left( { - \frac{1}{2}\cos 2x} \right)\left( {2xdx} \right)} \cr & \int {{x^2}\sin 2xdx} = - \frac{1}{2}{x^2}\cos 2x + \int {x\cos 2x} dx \cr & \cr & {\text{Integrating }}\int {x\cos 2x} dx \cr & {\text{Let }}u = x,\,\,\,\,dv = \cos 2xdx \cr & \,\,\,\,\,\,du = dx\,\,\,\,\,v = \frac{1}{2}\sin 2x \cr & {\text{Integrating}} \cr & \int {{x^2}\sin 2xdx} = - \frac{1}{2}{x^2}\cos 2x + \left( {\frac{1}{2}x\sin 2x - \int {\left( {\frac{1}{2}\sin 2x} \right)dx} } \right) \cr & \int {{x^2}\sin 2xdx} = - \frac{1}{2}{x^2}\cos 2x + \frac{1}{2}x\sin 2x - \frac{1}{2}\int {\sin 2xdx} \cr & \int {{x^2}\sin 2xdx} = - \frac{1}{2}{x^2}\cos 2x + \frac{1}{2}x\sin 2x - \frac{1}{4}\cos 2x + C \cr & {\text{Factoring}} \cr & \int {{x^2}\sin 2xdx} = \frac{1}{4}\left( {-1 - 2{x^2}} \right)\cos 2x + \frac{1}{2}x\sin 2x + C \cr} $$
